Calculating Volume and Side Length of a Cut Cone - Is This Right

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SUMMARY

The discussion centers on calculating the volume and slant height of a truncated cone, specifically a container with a top radius of 10 cm and a bottom radius of 20 cm, yielding a volume of 500 cm³. The user derived the height (h) as 1.5915 cm and calculated the slant height (s) using the Pythagorean theorem, resulting in s = 10.4944 cm. The volume formula for the truncated cone is provided as V = (πh/3)(R² + rR + r²), where R is the larger radius, r is the smaller radius, and h is the height.

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Ryoukomaru
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This is a problem I had in a test and almost everyone got different answers for it, we discussed and well, I spotted mistakes in their solutions so I think mine is right but I wanted to check here and also ask if there is an easier/faster way to do it.

There is a container that is similar to the bottom part of a cone which is cut into half. Top radius is [tex]10cm[/tex] and bottom is [tex]20cm[/tex]. And the volume of this container is [tex]500cm^{3}[/tex]. What is the length of the slanted side ?

So what I did was, first I drew this. ;P (See attachment)

Then what I did was to write an equation for [tex]V_2[/tex] in terms of [tex]V_1[/tex] and [tex]V_T[/tex]

By solving the equation for height, I got [tex]h=1.5915[/tex]
Then I used Pythagoras' theorem to find the length of the slanted side and it comes to:
[tex]10^2+2h^2=s^2=>100+3.183^2=s^2=> s=10.4944[/tex]

Is this correct ? And is there a formula to find the volume of this shape ?
 

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The volume of the "flower pot" section is

[tex]V = \frac {\pi h}{3} (R^2 + rR + r^2)[/tex]

where R is the large radius, r is the small radius and h is the same as your x. It's simply the difference between the volume of the large cone and the small cone.
 

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